您如何擺脫多余的強制轉換到已實現的接口？

Question

假設我有一個界面

public interface ICardSuit {
    /**short name*/
    public String getName();

    /** the colour of this card*/
    public ICardColour getColour();
}

我決定用一個枚舉實現：

public enum  CardSuit implements ICardSuit {
    HEART{
        @Override
        public ICardColour getColour() {
            return CardColour.RED;
        }
    },
    SPADE{
        @Override
        public ICardColour getColour() {
            return CardColour.BLACK;
        }
    },
    DIAMOND{
        @Override
        public ICardColour getColour() {
            return CardColour.RED;
        }
    },
    CLUBS {
        @Override
        public ICardColour getColour() {
            return CardColour.BLACK;
        }
    }
    ;

    @Override
    public String getName() {
        return this.name();
    }
}

現在，我想對其進行測試（使用kotlintest，因為我對此很喜歡）：

class CardSuitTest : FunSpec(){
    init {
        test("there are exactly four suits"){CardSuit.values().size shouldBe 4}
        test("suits implement interface"){CardSuit.values().forEach { it shouldBe instanceOf(ICardSuit::class) }}
        test("suits have correct names"){
            val suits = CardSuit.values() as Array<out ICardSuit>
            suits.forEach { when(it.name){
                "HEART" -> it should beTheSameInstanceAs(CardSuit.HEART as ICardSuit)
                "SPADE" -> it should beTheSameInstanceAs(CardSuit.SPADE as ICardSuit)
                "DIAMOND" -> it should beTheSameInstanceAs(CardSuit.DIAMOND as ICardSuit)
                "CLUBS" -> it should beTheSameInstanceAs(CardSuit.CLUBS as ICardSuit)
            } }
        }
        test("suits have correct colours"){
            CardSuit.values().forEach { when(it){
                CardSuit.HEART,CardSuit.DIAMOND -> it.colour shouldBe CardColour.RED
                CardSuit.CLUBS, CardSuit.SPADE -> it.colour shouldBe CardColour.BLACK
            } }
        }
    }
}

我需要轉換為ICardSuit因為如果我不這樣做，編譯器會抱怨

None of the following functions can be called with the arguments supplied.

* T.should(Matcher<T>)   where T cannot be inferred for    infix fun <T> T.should(matcher: Matcher<T>): Unit defined in io.kotlintest.matchers

* ICardSuit.should((ICardSuit) → Unit)   where T = ICardSuit for    infix fun <T> T.should(matcher: (T) → Unit): Unit defined in io.kotlintest.matchers

我想保留as Array<out ICardSuit>因為這是確保我僅訪問接口屬性的最簡單方法，

但是我真的不喜歡強制轉換要測試的實例。

有什么我可以做的嗎？

Answer 1

您是否需要使用匹配器beSameInstanceAs的特定原因？

您可以執行以下操作：

val suits = CardSuit.values() as Array<out ICardSuite>

suits.forEach {
    when (it.name) {
        "HEART" -> it shouldBe CardSuit.HEART
        "SPADE" -> it shouldBe CardSuit.SPADE
   }
}

但是，如果您確實要使用beSameInstanceAs ，則可以：

suits.forEach {
    when(it.name) {
       "HEART" -> it shouldBeSameInstanceAs CardSuit.HEART
       "SPADE" -> it shouldBeSameInstanceAs CardSuit.SPADE
    }
}

我在這里並沒有收到編譯器的任何投訴

Answer 2

beTheSameInstanceAs(CardSuit.HEART)返回Matcher<CardSuit> ，因此它不能匹配任意ICardSuit 。 這是有道理的（盡管Matcher可能是反變數的，在這里您需要協變）。 但是你可以：

1. 明確調用beTheSameInstanceAs<ICardSuit>(CardSuit.HEART) 。

2. 創建一個輔助函數

    inline fun <T1, reified T2 : T1> Matcher<T2>.widen() = object : Matcher<T1>() { override fun test(value: T1) = if (value is T2) this.test(value) else Result(false, "$value is not a ${T2::class.name}", "$value is a ${T2::class.name}") } 

   並打電話

    it should beTheSameInstanceAs(CardSuit.HEART).widen() 

   （我認為類型推斷應該在這里起作用）。

3. 因為beTheSameInstanceAs(x)實際上可以匹配任何東西，所以聲明一個等效函數，該函數返回Matcher<Any> ：

    fun beTheSameInstanceAsAny(x: Any) = beTheSameInstanceAs(x) // usage "HEART" -> it should beTheSameInstanceAsAny(CardSuit.HEART)